Lab8
Joshua Wiseman
Wednesday, March 29, 2023
Task 1
Working Directory
getwd()## [1] "/home/joshua/Desktop/MainFolder/OuClasses/Spring 2023/Applied Statistical Methods/FALL224753wise0046/LABS/LAB8"Task 2
runif(10,0,5)
sample = runif(10,0,5)
sample## [1] 0.4591017 3.7529155 1.9554246 4.9362817 1.8311409 3.3668557 3.2066818
## [8] 3.5099635 1.1258583 3.3858129Mean + Variance of uniform
# Mean
mu = (0+5)/2
mu## [1] 2.5# Variance
var = (5-0)^2/12
var## [1] 2.083333Mean + Variance of population
# Mean
x_hat = mean(sample)
x_hat## [1] 2.753004# Variance
s_squ = var(sample)
s_squ## [1] 1.857572Sum T
n = length(sample)
# Mean
T_mean = n * x_hat
T_mean## [1] 27.53004# Variance
T_var = n * s_squ
T_var## [1] 18.57572\(\bar{Y}\) mean
# Mean
Y_mean = mu
Y_mean## [1] 2.5# Variance
Y_var = s_squ / n
Y_var## [1] 0.1857572Explaination
myclt=function(n,iter){
y=runif(n*iter,0,5) # A
data=matrix(y,nr=n,nc=iter,byrow=TRUE) #B
sm=apply(data,2,sum) #C
hist(sm)
sm
}
w=myclt(n=10,iter=10000) #D
A is the assigning y with the a uniform distribution of sise (n) * iterations along with a value being 0 and b value being 5.
B is making a matrix with the uniform distribution making the number of rows n in length and the number of columns the number of iterations in length. Displayed by rows.
C is the applying the matrix with Margin = 2 and the sum resulting in getting the col.sums
D is calling the function myclt to have a length of n = 10 and iterations = 10000.
\(\bar{w} \wedge s^2_w\)
# Mean
mean(w)## [1] 24.99918# Variance
var(w)## [1] 21.1465Changed myclt
myclt=function(n,iter){
y=runif(n*iter,0,5) # A
data=matrix(y,nr=n,nc=iter,byrow=TRUE) #B
sm=apply(data,2,sum) #C
h=hist(sm,plot=FALSE)
hist(sm,col=rainbow(length(h$mids)),freq=FALSE,main="Distribution of the sum of uniforms")
curve(dnorm(x,mean=n*(5)/2,sd=sqrt(n*(5)^2/12)),add=TRUE,lwd=2,col="Blue")
sm
}
w=myclt(n=10,iter=10000) #D
Task 3
mycltu()
################### uniform ##########################
### CLT uniform
## my Central Limit Function
## Notice that I have assigned default values which can be changed when the function is called
mycltu=function(n,iter,a=0,b=10){
## r-random sample from the uniform
y=runif(n*iter,a,b)
## Place these numbers into a matrix
## The columns will correspond to the iteration and the rows will equal the sample size n
data=matrix(y,nr=n,nc=iter,byrow=TRUE)
## apply the function mean to the columns (2) of the matrix
## these are placed in a vector w
w=apply(data,2,mean)
## We will make a histogram of the values in w
## How high should we make y axis?
## All the values used to make a histogram are placed in param (nothing is plotted yet)
param=hist(w,plot=FALSE)
## Since the histogram will be a density plot we will find the max density
ymax=max(param$density)
## To be on the safe side we will add 10% more to this
ymax=1.1*ymax
## Now we can make the histogram
hist(w,freq=FALSE, ylim=c(0,ymax), main=paste("Histogram of sample mean",
"\n", "sample size= ",n,sep=""),xlab="Sample mean")
## add a density curve made from the sample distribution
lines(density(w),col="Blue",lwd=3) # add a density plot
## Add a theoretical normal curve
curve(dnorm(x,mean=(a+b)/2,sd=(b-a)/(sqrt(12*n))),add=TRUE,col="Red",lty=2,lwd=3) # add a theoretical curve
## Add the density from which the samples were taken
curve(dunif(x,a,b),add=TRUE,lwd=4)
}
mycltu(n=20,iter=100000)
Examine
w=apply(data,2,mean), how does the apply function use the 2?
apply the mean function to each column of matrix
How many terms are in w, when mycltu(n=20,iter=100000) is called?
20 * 100000## [1] 2e+06curve(dnorm(x,mean=(a+b)/2,
sd=(b-a)/(sqrt(12*n))),add=TRUE,col=“Red”,lty=2,lwd=3):
graph the curve of the distribution of the given mean and variance with the use of sd
Record Plots
mycltu(n=1,iter=10000)
mycltu(n=2,iter=10000)
mycltu(n=3,iter=10000)
mycltu(n=5,iter=10000)
mycltu(n=10,iter=10000)
mycltu(n=30,iter=10000)
I conclude that the function becomes more like a normal distribution the more iterations and the greater n is in size.
Task 4
mycltb()
############################## Binomial #########
## CLT Binomial
## CLT will work with discrete or continuous distributions
## my Central Limit Function
## Notice that I have assigned default values which can be changed when the function is called
mycltb=function(n,iter,p=0.5,...){
## r-random sample from the Binomial
y=rbinom(n*iter,size=n,prob=p)
## Place these numbers into a matrix
## The columns will correspond to the iteration and the rows will equal the sample size n
data=matrix(y,nr=n,nc=iter,byrow=TRUE)
## apply the function mean to the columns (2) of the matrix
## these are placed in a vector w
w=apply(data,2,mean)
## We will make a histogram of the values in w
## How high should we make y axis?
## All the values used to make a histogram are placed in param (nothing is plotted yet)
param=hist(w,plot=FALSE)
## Since the histogram will be a density plot we will find the max density
ymax=max(param$density)
## To be on the safe side we will add 10% more to this
ymax=1.1*ymax
## Now we can make the histogram
## freq=FALSE means take a density
hist(w,freq=FALSE, ylim=c(0,ymax),
main=paste("Histogram of sample mean","\n", "sample size= ",n,sep=""),
xlab="Sample mean",...)
## add a density curve made from the sample distribution
#lines(density(w),col="Blue",lwd=3) # add a density plot
## Add a theoretical normal curve
curve(dnorm(x,mean=n*p,sd=sqrt(p*(1-p))),add=TRUE,col="Red",lty=2,lwd=3)
}
mycltb(n=5,iter=10000,p=0.5)
Graphs with p=0.3
mycltb(n=4,iter=10000,p=0.3)
mycltb(n=5,iter=10000,p=0.3)
mycltb(n=10,iter=10000,p=0.3)
mycltb(n=20,iter=10000,p=0.3)
Graphs with p=0.7
mycltb(n=4,iter=10000,p=0.7)
mycltb(n=5,iter=10000,p=0.7)
mycltb(n=10,iter=10000,p=0.7)
mycltb(n=20,iter=10000,p=0.7)
Graphs with p=0.5
mycltb(n=4,iter=10000,p=0.5)
mycltb(n=5,iter=10000,p=0.5)
mycltb(n=10,iter=10000,p=0.5)
mycltb(n=20,iter=10000,p=0.5)
I conclude that the function becomes more like a normal distribution the more iterations and the greater n is in size like previously. However, because the p value is changing, after enough iterations it will skew towards the percentage or p value.